3.2.8 \(\int \csc ^2(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx\) [108]

Optimal. Leaf size=70 \[ \frac {2 F\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{b}-\frac {2 \cos (2 a+2 b x) \sqrt {\sin (2 a+2 b x)}}{b}+\frac {\csc ^2(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{b} \]

[Out]

-2*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticF(cos(a+1/4*Pi+b*x),2^(1/2))/b+csc(b*x+a)^2*sin(2*b*x
+2*a)^(5/2)/b-2*cos(2*b*x+2*a)*sin(2*b*x+2*a)^(1/2)/b

________________________________________________________________________________________

Rubi [A]
time = 0.04, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {4385, 2715, 2720} \begin {gather*} \frac {2 F\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{b}-\frac {2 \sqrt {\sin (2 a+2 b x)} \cos (2 a+2 b x)}{b}+\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \csc ^2(a+b x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^2*Sin[2*a + 2*b*x]^(3/2),x]

[Out]

(2*EllipticF[a - Pi/4 + b*x, 2])/b - (2*Cos[2*a + 2*b*x]*Sqrt[Sin[2*a + 2*b*x]])/b + (Csc[a + b*x]^2*Sin[2*a +
2*b*x]^(5/2))/b

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 4385

Int[((e_.)*sin[(a_.) + (b_.)*(x_)])^(m_)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(e*Sin[a + b*
x])^m*((g*Sin[c + d*x])^(p + 1)/(2*b*g*(m + p + 1))), x] + Dist[(m + 2*p + 2)/(e^2*(m + p + 1)), Int[(e*Sin[a
+ b*x])^(m + 2)*(g*Sin[c + d*x])^p, x], x] /; FreeQ[{a, b, c, d, e, g, p}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b,
2] &&  !IntegerQ[p] && LtQ[m, -1] && NeQ[m + 2*p + 2, 0] && NeQ[m + p + 1, 0] && IntegersQ[2*m, 2*p]

Rubi steps

\begin {align*} \int \csc ^2(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx &=\frac {\csc ^2(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{b}+6 \int \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx\\ &=-\frac {2 \cos (2 a+2 b x) \sqrt {\sin (2 a+2 b x)}}{b}+\frac {\csc ^2(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{b}+2 \int \frac {1}{\sqrt {\sin (2 a+2 b x)}} \, dx\\ &=\frac {2 F\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{b}-\frac {2 \cos (2 a+2 b x) \sqrt {\sin (2 a+2 b x)}}{b}+\frac {\csc ^2(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{b}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.93, size = 73, normalized size = 1.04 \begin {gather*} \frac {2 \sqrt {\sin (2 (a+b x))}-\frac {\sqrt {2} F\left (\text {ArcSin}(\cos (a+b x)-\sin (a+b x))\left |\frac {1}{2}\right .\right ) (\cos (a+b x)+\sin (a+b x))}{\sqrt {1+\sin (2 (a+b x))}}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^2*Sin[2*a + 2*b*x]^(3/2),x]

[Out]

(2*Sqrt[Sin[2*(a + b*x)]] - (Sqrt[2]*EllipticF[ArcSin[Cos[a + b*x] - Sin[a + b*x]], 1/2]*(Cos[a + b*x] + Sin[a
+ b*x]))/Sqrt[1 + Sin[2*(a + b*x)]])/b

________________________________________________________________________________________

Maple [A]
time = 7.87, size = 111, normalized size = 1.59

method	result	size
default	\(\frac {\sqrt {2}\, \left (\sqrt {2}\, \left (\sqrt {\sin }\left (2 x b +2 a \right )\right )+\frac {\sqrt {2}\, \sqrt {\sin \left (2 x b +2 a \right )+1}\, \sqrt {-2 \sin \left (2 x b +2 a \right )+2}\, \sqrt {-\sin \left (2 x b +2 a \right )}\, \EllipticF \left (\sqrt {\sin \left (2 x b +2 a \right )+1}, \frac {\sqrt {2}}{2}\right )}{2 \cos \left (2 x b +2 a \right ) \sqrt {\sin \left (2 x b +2 a \right )}}\right )}{b}\)	\(111\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)^2*sin(2*b*x+2*a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

2^(1/2)*(2^(1/2)*sin(2*b*x+2*a)^(1/2)+1/2*2^(1/2)*(sin(2*b*x+2*a)+1)^(1/2)*(-2*sin(2*b*x+2*a)+2)^(1/2)*(-sin(2
*b*x+2*a))^(1/2)*EllipticF((sin(2*b*x+2*a)+1)^(1/2),1/2*2^(1/2))/cos(2*b*x+2*a)/sin(2*b*x+2*a)^(1/2))/b

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^2*sin(2*b*x+2*a)^(3/2),x, algorithm="maxima")

[Out]

integrate(csc(b*x + a)^2*sin(2*b*x + 2*a)^(3/2), x)

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^2*sin(2*b*x+2*a)^(3/2),x, algorithm="fricas")

[Out]

integral(csc(b*x + a)^2*sin(2*b*x + 2*a)^(3/2), x)

________________________________________________________________________________________

Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)**2*sin(2*b*x+2*a)**(3/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3003 deep

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^2*sin(2*b*x+2*a)^(3/2),x, algorithm="giac")

[Out]

integrate(csc(b*x + a)^2*sin(2*b*x + 2*a)^(3/2), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\sin \left (2\,a+2\,b\,x\right )}^{3/2}}{{\sin \left (a+b\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(2*a + 2*b*x)^(3/2)/sin(a + b*x)^2,x)

[Out]

int(sin(2*a + 2*b*x)^(3/2)/sin(a + b*x)^2, x)

________________________________________________________________________________________